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Trying to understand relativity of lengths and times

  1. Oct 30, 2006 #1
    let K' = a stationary rigid rod. let its length be L (the length between the two ends A and B of K' using a measuring-rod). the axis of K' is lying along the x-axis of a stationary system of coordinates K.

    let K be provided with clocks which synchronize.

    let a constant speed v in the direction of increasing x be imparted to K'.

    let rAB = the distance (as measured by the measuring-rod already employed to measure L) between two points on the x-axis of K where the two ends A and B of K’ are located at a definite time.

    let clocks be placed at the two ends A and B of K’, and let these clocks synchronize with the clocks of K

    let a ray of light depart from A at tA in the direction of B, and let the ray of light be reflected at B at tB and reach A again at t’A.

    time here denotes time of K and also position of hands of the moving clock at the place under discussion.

    observers moving with K' would declare that (tB - tA) = rAB/(c - v) is not equal to (t’A - tB) = rAB/(c + v).

    while observers in K would declare that (tB - tA) = rAB/(c - v) = (t’A - tB) = rAB/(c + v).

    if L = 299,792,458 meters, and v = 0.5*c, what is the value of (tB - tA) and (t’A - tB) using the lorentz transformations if the observers are moving with K'?

    and if L = 299,792,458 meters, and v = 0.5*c, what is the value of (tB - tA) and (t’A - tB) using the lorentz transformations if the observers are in K?

    thanks. i think the answer is going to help me understand the lorentz transformations and how to apply them in order to understand relativity of lengths and times. thanks again.
     
    Last edited: Oct 30, 2006
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  3. Oct 30, 2006 #2

    Doc Al

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    I find it extremely difficult to understand your posts. Let me try my own version of what you might be getting at:

    A stick, proper length L, has ends A and B. There are clocks at A and B attached to the stick. These clocks are synchronized in their own frame.

    A beam of light flashes at A, travels to B, then is reflected back to A.

    Realize that different observers will measure different times and locations for these events (the flash, the reflection, the return).

    In the frame of the stick (let's call that the "moving frame"), the light travels a distance L in time T. c = L/T.

    In a frame in which the stick is seen to move at speed v (let's call that the "rest frame"), the light takes time T' to get from A to B. The distance covered is L' + vT', where L' is the length of the stick according to the rest frame and vT' is the distance that end B moves in that time according to the rest frame.

    Now, the time for the round trip of the light--according to the moving frame--is just 2T. That time will be recorded on the clock at point A.

    Now, according to the rest frame, the time for round trip of the light will be 2T/sqrt(1 - v^2/c^2).

    With this as a start, can you rephrase your question?
     
  4. Oct 30, 2006 #3

    A stick, proper length L, has ends A and B. There are clocks at A and B attached to the stick. These clocks are synchronized in their own frame.

    A beam of light flashes at A at TA (the position of the hands of the clock at A), travels to B at TB (the position of the hands of the clock at B), then is reflected back to A at T‘A (the position of the hands of the clock at A).

    In the frame of the stick (let's call that the "moving frame"), the light travels from A to B a distance L in time TB - TA. c = L/(TB - TA). And in the moving frame, the light travels from B to A a distance L in time T‘A - TB. c = L/(T’A - TB).

    In a frame in which the stick is seen to move at speed v (let's call that the "rest frame"), the beam of light flashes at A at TA, travels to B at TB, then is reflected back to A at T’A.

    Realize that the clocks at A and B were adjusted before the stick began to move at speed v in the rest frame in order that they synchronize with the clocks in the rest frame when the stick is then seen to move at speed v in the rest frame.

    In the rest frame, the light travels from A to B a distance L’ (where L' is the length of the stick according to the rest frame) in time TB - TA. c = L’/(TB - TA) + v. And in the rest frame, the light travels from B to A in time T’A - TB. c = L’/(T’A - TB) - v. TB - TA = T’A - TB.

    In the moving frame, c = L’/(TB - TA) + v and c = L’/(T’A - TB) - v. (TB - TA) is not equal to (T’A - TB).

    what is the value of (TB - TA), (T'A - TB), and L' in the rest frame and the moving frame respectively if L = 299,792,458 meters? thanks.
     
    Last edited: Oct 30, 2006
  5. Oct 30, 2006 #4

    JesseM

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    You mean the clocks A and B are now in sync in the frame where the stick moves at velocity v, and out-of-sync in the rest frame of the stick? But do you also mean that the rates of the clocks at A and B are adjusted so that they appear to tick at the correct rate in the "rest frame" where the stick is moving at speed v? In other words, the two clocks have been artificially sped up to tick at [tex]1/\sqrt{1 - v^2/c^2}[/tex] the normal rate in their own rest frame?
    "Rest frame" refers to the frame where the stick is moving, right? Your post would be more clear if instead of "rest frame" and "moving frame" you specified whose rest frame you were talking about, as in "the rest frame of the stick" or "the rest frame of the observer who sees the stick moving at velocity v". But assuming you mean the latter, then if the clocks at A and B have been adjusted so that they are not only synchronized in the observer's rest frame, but also ticking at the correct rate in the observer's frame, then these equations will be correct, because in that case (TB - TA) will be the time the observer measures between the light passing A and the light passing B, and during that time the ruler will have moved forward by a distance of v*(TB - TA), so the total distance the light will have covered between those events is L' + v*(TB - TA) in this frame, and the total time between the events is (TB - TA) in this frame, and distance/time in the observer's frame must be c. Likewise, on the way back the light will only have to cover a distance of L' - v*(T'A - TB) because A is moving towards the point where the light was reflected, so your second equation is also right.
    Why would you say "in the moving frame" when you are using clocks synchronized in the observer's rest frame (and, if I'm understanding right, also running at the correct rate in the observer's rest frame), and the length L' seen in the observer's rest frame? When you say "in frame x" you normally mean "using frame x's definition of the distance between the events and frame x's definition of the time between them".
    If the clocks at A and B are synchronized in the frame of the observer who sees the ruler moving at velocity v, that's correct.
    Assuming (TB - TA) is simply the time for the light to go from A to B in the frame of the observer who sees the ruler moving at v, and L' is the length of the ruler in that observer's frame, then L' is easy to find, it's just [tex]299,792,458*\sqrt{1 - v^2/c^2}[/tex] due to length contraction. To find (TB - TA) and (T'A - TB) you'd have to also know v, but it would just be a matter of solving the equations you already gave, c = L'/(TB - TA) + v (which can be rearranged to give (TB - TA) = L'/(c - v)) and c = L'/(T'A - TB) - v (which can be rearranged to give (T'A - TB) = L'/(c + v)).
     
  6. Oct 30, 2006 #5

    Doc Al

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    So far, so good.

    ah... here's a problem. By starting with the stick "at rest" and then having it move at speed "v" you are introducing an acceleration and its associated complications. In order to preserve the synchonization of clocks A and B as the stick is accelerated, you would have to accelerate the ends of the stick (and all points in between) uniformly with respect to the "rest frame". The stick will break (or at least be forced to stretch) under the stress of that operation--it will no longer have a proper length L. Even if you succeed in getting the stick to move at speed v without disintegrating, the clocks will still be running slowly compared to clocks in the rest frame.

    I strongly advise not introducing such accelerations when first trying to sort out the basics of special relativity. Instead, imagine the stick to have been always in motion at speed v with respect to the "rest frame". Pose your questions with respect to that model and it'll be much easier to see how the Lorentz transformations work.
     
  7. Oct 30, 2006 #6

    i understand. let me try it this way (i am quoting A. Enstein in the process, pages 40-42 in The Principle of Relativity):

    A stick, proper length L, has ends A and B.

    In a frame in which the stick is seen to move at speed v (let's call that the "rest frame"), there are clocks set up on the x-axis of this "rest frame". These clocks are synchronized. L' is the distance between two points on the x-axis of this "rest frame" where the two ends A and B of the stick are located at a definite time.

    There are clocks attached at the two ends A and B of the stick which synchronize with the clocks set up on the x-axis of the "rest frame", that is to say that their indications correspond at any instant to the "time of the 'rest frame'" at the places where they happened to be. These clocks are therefore, synchronous in the "rest frame".

    A beam of light flashes at A at the time* TA (the position of the hands of the clock at A), travels to B at the time TB (the position of the hands of the clock at B), then is reflected back to A at the time T‘A (the position of the hands of the clock at A). The two clocks synchronize if

    TB - TA = T'A - TB

    for simplicity, let us assume that the moving clocks and the clocks in the "rest frame" simultaneously mark TA = 0 s.

    *"time" here denotes "time of the 'rest frame'" and also "position of hands of the moving clock situated at the place under discussion."

    Taking into consideration the principle of the constancy of the velocity of light we find that

    TB - TA = L'/(c - v) and T'A - TB = L'/(c + v)

    observers moving with the stick would find that the two clocks were not synchronous, while observers in the "rest frame" would declare the clocks to be synchronous.

    using the lorentz transformations, find the value of (TB - TA) and L' if L = 299,792,458 meters and v = 0.5*c, such that the observers moving with the stick would find that the two clocks were not synchronous, while the observers in the "rest frame" would declare the clocks to be synchronous. thanks!
     
    Last edited: Oct 31, 2006
  8. Oct 31, 2006 #7

    JesseM

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    These are the formulas I gave at the end of my last post.
    As I mentioned above, this scenario is also assuming the clocks are ticking at the correct rate in the "rest frame", which means that in the frame of the stick the clocks would appear artificially sped up by a factor of [tex]1/\sqrt{1 - v^2/c^2}[/tex].
    Just plug into the formulas, using the length contraction formula to find L' given L. In this case L' = 299,792,458*[tex]\sqrt{1 - 0.5^2}[/tex] meters, so TB - TA = (299,792,458 meters)*[tex]\sqrt{1 - 0.5^2}[/tex]/(1c - 0.5c) = (299,792,458 meters)*[tex]\sqrt{1 - 0.5^2}[/tex]/(299,792,458 meters/second)*(0.5) = [tex](\sqrt{1 - 0.5^2})/0.5[/tex]*(1 second) = 1.73205081 seconds. Similarly, T'A - TB = (299,792,458 meters)*[tex]\sqrt{1 - 0.5^2}[/tex]/(1c + 0.5c) = (299,792,458 meters)*[tex]\sqrt{1 - 0.5^2}[/tex]/(299,792,458 meters/second)*(1.5) = [tex](\sqrt{1 - 0.5^2})/1.5[/tex]*(1 second) = 0.577350269 seconds.
     
  9. Oct 31, 2006 #8
    v = 0.5c
    gamma = 1.1547

    Rest Frame …

    TB-TA = T(out) = 1 sec = 1/2(2L/c)
    T'A-TB = T(back) = 1 sec = 1/2(2L/c)
    L = 1 lt-sec = 299,792,458 meters (proper length)

    Moving frame …

    TB-TA = T(out) = 1.732 sec = (L/gamma)[1/(c-v)]
    T'A-TB = T(back) = 0.57733 sec = (L/gamma)[1/(c+v)]
    L’ = 0.866 lt-sec = 259,620,268 meters = L/gamma (contracted length)

    By the Lorentz Transformations ...

    Per Rest Frame (by definition) ...

    TB-TA = 1 sec = 1/2(2L/c)
    T'A-TB = 1 sec = 1/2(2L/c)
    L = 1 lt-sec = 299,792,458 meters

    Per Moving Frame (per LTs) ...

    t(out) = gamma(T+vX/c2) = TB-TA

    t(out) = gamma(T+vX/c2)
    t(out) = (1.1547)[(1)+(0.5)(1)/(1)2]
    t(out) = 1.732 sec​

    t(back) = gamma(T-vX/c2) = T'A-TB

    t(back) = gamma(T-vX/c2)
    t(back) = (1.1547)[(1)-(0.5)(1)/(1)2]
    t(back) = 0.5774 sec​

    Verification ... Roundtrip

    t(out)+t(back) = gamma[T(out)+T(back)]
    1.732 sec + 0.5774 sec = (1.1547)[1 sec + 1sec]
    2.3094 sec = 2.3094 sec​

    x(out) = gamma(X-vT)

    x(out) = gamma(X-vT)
    x(out) = (1.1547)[(1)-(0.5c)/(1)2]
    x(out) = (1.1547)[1-0.5]
    x(out) = 1.732 lt-sec​

    x(back) = gamma(X+vT)

    x(back) = gamma(X+vT)
    x(back) = (1.1547)[(1)+(0.5c)/(1)2]
    x(back) = (1.1547)[1+0.5]
    x(back) = 0.5774 lt-sec​

    L' = x’ = (x-vt) ... calculated by the outbound leg

    x’ = (x-vt)
    x’ = ct-vt
    x’ = t(c-v)
    x’ = (1.732 sec)(1-0.5)
    x’ = (1.732 sec)0.5
    x’ = 0.866 lt-sec
    L' = x’ = 0.866 lt-sec = L/gamma

    L' = x’ = (x+vt) ... calculated by the return leg

    x’ = (x+vt)
    x’ = ct+vt
    x’ = t(c+v)
    x’ = (0.5774 sec)(1+0.5)
    x’ = (0.5774 sec)1.5
    x’ = 0.866 lt-sec
    L' = x’ = 0.866 lt-sec = L/gamma

    pess
     
  10. Oct 31, 2006 #9

    JesseM

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    myoho.renge.kyo is actually defining "rest frame" and "moving frame" in the opposite way--the "moving frame" is the rest frame of the stick (where its length is 1 light-second), and the "rest frame" is the frame where the stick is moving at velocity v. Also, TB - TA and T'A - TB are defined to be the time intervals in the "rest frame" where the stick is moving, not the time in the stick's own frame. This sort of confusion is why I made this suggestion earlier:
    Aside from that, your numbers for the times agree with the ones I got in my last post.
     
  11. Oct 31, 2006 #10
    you say "Rest Frame..."

    what do you mean?
     
    Last edited: Oct 31, 2006
  12. Oct 31, 2006 #11
    the "rest frame" is the frame in which the stick is seen to move at speed v.

    the "moving frame" is the stick that moves at speed v in the "rest frame".
     
  13. Oct 31, 2006 #12
    are those values in the "rest frame" or in the stick?

    what are the values in the other frame (the "rest frame" or the stick)?

    i appreciate your reply. thank you!
     
  14. Oct 31, 2006 #13
    according to your calculations, would the observers moving with the moving stick find that the two clocks were not synchronous? and at the same time would the observers in the "rest frame" declare the clocks to be synchronous?
     
    Last edited: Oct 31, 2006
  15. Oct 31, 2006 #14

    JesseM

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    Those were the values in the "rest frame", ie the frame where the stick was moving at velocity v. As I explained earlier, in this frame the time between the light passing A and the light passing B is (TB - TA) (because you defined the clocks at the end of the ruler to be synchronized in this frame and ticking at the correct rate for this frame, meaning they'd be out-of-sync and ticking too fast in the stick's own rest frame), the length of the stick is L' in this frame, and because the stick has moved forward a distance of v*(TB - TA) in the time (TB - TA), the total distance the light covers between passing A and passing B is L' + v*(TB - TA) in this frame. So, (distance as measured in 'rest frame')/(time as measured in 'rest frame') is equal to [L' + v*(TB - TA)]/(TB - TA), or L'/(TB - TA) + v. And of course, distance/time for a light beam in any inertial frame must be c, so c = L'/(TB - TA) + v, which you can rearrange to get (TB - TA) = L'/(c - v).

    In the stick's own frame (what you called the 'moving frame'), the stick has length L, and since the stick is at rest this is the distance the light will cover in both directions, and of course (distance in moving frame)/(time in moving frame) must equal c, so the light will take a time of L/c to get from A to B, and likewise it will take L/c to get from B to A, meaning the times will be 1 second in either direction.
     
    Last edited: Oct 31, 2006
  16. Oct 31, 2006 #15
    i was thinking more along these lines:

    given:

    L = 299,792,458 m

    v = 0.5*c

    1/sqrt(1 - v^2/c^2) = 1.1547

    the moving clock at A and the clock in the "rest frame" mark TA = 0 s.

    -------------------------------

    let the moving clocks measure TB - TA = T = 1 s (299,792,458 m/c).

    let the clocks in the "rest frame" measure TB - TA = t

    ------------------------------

    by the lorentz transformations,

    T = (t - v*L'/c^2)/1.1547

    1.1547*T = t - v*L'/c^2

    1.1547*T = t - (0.5*c)*L'/c^2

    1.1547*T = t - 0.5*L'/c

    t = 1.1547*T + 0.5*L'/c

    ------------------------------

    L = (L' - v*t)/1.1547

    1.1547*L = L' - v*t

    1.1547*L = L' - 0.5*c*t

    0.5*c*t = L' - 1.1547*L

    t = (L' - 1.1547*L)/0.5*c

    --------------------------------

    1.1547*T + 0.5*L'/c = (L' - 1.1547*L)/0.5*c

    1.1547*T*(0.5*c) + 0.25*L' = L' - 1.1547*L

    L' - 0.25*L' = 1.1547*T*(0.5*c) + 1.1547*L

    L'*(1 - 0.25) = 1.1547*(0.5*T*c + L)

    L' = 1.1547*(0.5*T*c + L)/0.75

    L' = 1.1547*(149,896,229 m + 299,792,458 m)/0.75

    L' = 1.1547*449,688,687 m/0.75

    L' = 692,340,702.505 m

    ---------------------------------

    t = 1.1547*T + 0.5*L'/c

    t = 1.1547 s + 1.1547 s

    t = 2.3094 s

    ----------------------------------

    t = (L' - 1.1547*L)/0.5*c

    t = (692,340,702.505 m - 346,170,351.253 m)/(149,896,229 m/s)

    t = 2.3094 s

    --------------------------------------

    692,340,702.505 m/(c - v) = 4.6188 s = T

    692,340,702.505 m/(c + v) = 1.5396 s = (T'A - T)

    we can see that the observers moving with the stick find that the two clocks were not synchronous.

    -------------------------------------------

    692,340,702.505 m/(c - v) = 2.3094 s = t (realize that v = 0)

    692,340,702.505 m/(c + v) = 2.3094 s = (T'A - t) (again realize that v = 0). the mark of T'A by the moving clock at A is 6.1584 s (4.6188 s + 1.5396 s), and the mark of T'A by the clock in the "rest frame" is 4.6188 s (2*2.3094 s).

    we can see here that at the same time, the observers in the "rest frame" declare the clocks to be synchronous.

    --------------------------------

    now i know what A. Einstein meant on p.42 in The Principle of Relativity (thanks to all of you!!).
     
    Last edited: Oct 31, 2006
  17. Oct 31, 2006 #16
    Well, I ran the calculation as though the stick were 1 light-sec long in the stick's proper frame. However, any way you slice it, clocks in sync on the ground (ie rest frame) appear unsynchronized per the moving stick. Clock is sync on the moving stick appear unsynchronized per the ground observers. The offset between clocks stems from the -vx/c^2 portion of the LT, so the offset is gamma(xv/c^2) where x is the contracted seperation in clocks per the observer.

    pess
     
  18. Oct 31, 2006 #17
    Indeed, I mistakenly assumed the stick's proper frame as the rest frame. Didn't read your secanrio setup carefully enough.

    pess
     
  19. Oct 31, 2006 #18
    yes, I misread the scenario setup. Thanx.

    I find it best to refer to the "stick frame" and the "observing frame". Either can be stationary per its own vantage, and either considers itself at rest per its own perspective. Two bodies can be at rest with each other when in inertial motion. But then, all inertial bodies are at rest.

    pess
     
  20. Oct 31, 2006 #19

    JesseM

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    I don't know what event you're trying to transform the coordinates of here--why do you have L' in there? In the "rest frame", the position-coordinate of the clock at B will only be L' when t=0, at later times that end of the ruler will have moved. Also, you shouldn't be dividing by 1.1547, you should be multiplying by it--the Lorentz transform equation for time is [tex]T = \gamma*(t - vx/c^2)[/tex], with [tex]\gamma = 1/\sqrt{1 - v^2/c^2}[/tex], which in this case would be [tex]1/\sqrt{1 - 0.5^2} = 1.1547[/tex].

    In the "rest frame", the A end of the stick was at position x=0 at time t=0, and the B end of the stick was at position x=L'. Because the stick is moving, the B end's position as a function of time will be x(t) = L' + vt. Meanwhile, the light beam's position as a function of time is x(t) = ct, so the light beam catches up with the B end when L' + vt = ct, or when t = L'/(c - v). With L=1 light-second and v=0.5c (and L' = L*0.866 = 0.866 light-seconds), this works out to t=1.732 seconds. Since this is along the path of the light beam, the position of this event must be x=1.732 light-seconds. You can then plug that into the Lorentz transform T = 1.1547*(1.732 s - (0.5 ls/s *1.732 ls)/(1 ls^2/s^2)) = 1.1547*(1.732 s - 0.5*1.732 s) = 1 s.

    Alternately, you could figure out the coordinates of the event of the light reaching B in the "moving frame" where the stick is at rest--in this frame, B is always at position X=L, so naturally the light beam reaches B at T=L/c. Then just plug that into the reversed version of the Lorentz transform:

    [tex]x = \gamma * (X + vT)[/tex]
    [tex]t = \gamma * (T + vX/c^2)[/tex]

    With X=L and T=L/c, this gives t=(1.1547)*(1.5L/c) = 1.732 seconds, and x=(1.1547)*(1.5L) = 1.732 light-seconds, same as above.

    Anyway, the rest of your post is wrong because your initial equation doesn't make sense, but if you think about precisely what event you want to transform, and what coordinates this event should have in either coordinate system, then you should be able to use the Lorentz transform to find the coordinates of the same event in the other coordinate system.
     
    Last edited: Oct 31, 2006
  21. Oct 31, 2006 #20

    JesseM

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    When L' is "ascertained" is irrelevant, L' is just the length of the stick in the "rest frame", which is the same at all times in that frame. What's important is that the stick is moving in the rest frame, so that even though it has a constant length L', the position of the end of the stick is changing over time. So, the position of the B-end at the moment the light reaches it is different than the position the B-end was at when the light was passing the A-end.

    Suppose I walk from one end of a train car to another, and the train car is moving down the tracks at high speed in your frame. Wouldn't you agree that the distance between my position when I was at the back end of the car and my position when I reached the front end of the car would be much larger than the length of the car itself in your frame, since the train car would have been moving forward while I was walking along it? It's exactly the same principle with the light beam moving from one end of the stick to the other in a frame where the stick is moving, the distance the light covers in that frame will not be equal to the length of the stick in that frame.
     
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